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# 数学代写|交换代数代写Commutative Algebra代考|MAT4200

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## 数学代写|交换代数代写Commutative Algebra代考|Basic equivalences

Although we will have no use for them in the sequel of these notes, in both commutative and (especially) homological algebra there is an important class of modules “dual” to the projective modules. They are characterized as follows.

Proposition 3.19. For a module $E$ over a ring $R$, the following are equivalent: (ii) If $\iota: M \rightarrow N$ is an injective $R$-module homomorphism and $\varphi: M \rightarrow E$ is any homomorphism, there exists at least one extension of $\varphi$ to a homomorphism $\Phi: N \rightarrow E$.
(iii) If $M \hookrightarrow N$, the natural map $\operatorname{Hom}(N, E) \rightarrow \operatorname{Hom}(M, E)$ is surjective.
(iv) The (contravariant) functor $\operatorname{Hom}(, E)$ is exact.
(v) Any short exact sequence of R-modules
$$0 \rightarrow E \stackrel{\iota}{\rightarrow} M \rightarrow N \rightarrow 0$$

splits: there exists an $R$-module map $\pi: M \rightarrow E$ such that $\pi \circ \iota=1_E$ and thus an internal direct sum decomposition $M=\iota(E) \oplus \operatorname{ker}(\pi) \cong E \oplus N$. A module satisfying these equivalent conditions is called injective.
EXERCISE 3.33. Prove Proposition 3.19.
EXERCISE 3.34. Show: an $R$-module $E$ is injective iff whenever $E$ is a submodule of a module $M, E$ is a direct summand of $M$.

Notice that the set of equivalent conditions starts with (ii)! This is to facilitate direct comparison to Proposition 3.10 on projective modules. Indeed, one should check that each of the properties (ii) through (v) are duals of the corresponding properties for projective modules: i.e., they are obtained by reversing all arrows. The difficulty here with property (i) is that if one literally reverses the arrows in the definition of free $R$-module to arrive at a “cofree” $R$-module, one gets a definition which is unhelpfully strong: the “cofree $R$-module on a set $X$ ” does not exist when $# X>1$ ! This can be remedied by giving a more refined definition of cofree module. For the sake of curiosity, we will give it later on in the exercises, but to the best of my knowledge, cofree $R$-modules by any definition do not play the fundamental role that free $R$-modules do.

## 数学代写|交换代数代写Commutative Algebra代考|Baer’s Criterion

ThEorem 3.20. (Baer’s Criterion $[\mathbf{B a 4 0}]$ ) For a module $E$ over a ring $R$, the following are equivalent:
(i) $E$ is injective.
(ii) For every ideal nonzero $I$ of $R$, every $R$-module map $\varphi: I \rightarrow E$ extends to an $R$-module map $\Phi: R \rightarrow E$.

Proof. (i) $\Longrightarrow$ (ii): this is a special case of condition (ii) of Proposition 3.19: take $M=I, N=R$.
(ii) $\Longrightarrow$ (i): Let $M$ be an $R$-submodule of $N$ and $\varphi: M \rightarrow E$ an $R$-module map. We need to show that $\varphi$ may be extended to $N$. Now the set $\mathcal{P}$ of pairs $\left(N^{\prime}, \varphi^{\prime}\right)$ with $M \subset N^{\prime} \subset N$ and $\varphi: N^{\prime} \rightarrow E$ a map extending $\varphi$ is nonempty and has an evident partial ordering, with respect to which the union of any chain of elements in $\mathcal{P}$ is again an element of $\mathcal{P}$. So by Zorn’s Lemma, there is a maximal element $\varphi^{\prime}: N^{\prime} \rightarrow E$. Our task is to show that $N^{\prime}=N$.
Assume not, and choose $x \in N \backslash N^{\prime}$. Put
$$I=\left(N^{\prime}: x\right)=\left{r \in R \mid r x \subset N^{\prime}\right}$$

one checks immediately that $I$ is an ideal of $R$ (a generalization to modules of the colon ideal we have encountered before). Consider the composite map
$$I \stackrel{\cdot x}{\rightarrow} N^{\prime} \stackrel{\varphi}{\rightarrow} E ;$$
by our hypothesis, this extends to a map $\psi: R \rightarrow E$. Now put $N^{\prime \prime}=\left\langle N^{\prime}, x\right\rangle$ and define $^4 \varphi^{\prime \prime}: N^{\prime \prime} \rightarrow E$ by
$$\varphi^{\prime \prime}\left(x^{\prime}+r x\right)=\varphi^{\prime}\left(x^{\prime}\right)+\psi(r) .$$
Thus $\varphi^{\prime \prime}$ is an extension of $\varphi^{\prime}$ to a strictly larger submodule of $N$ than $N^{\prime}$, contradicting maximality.

## 数学代写|交换代数代写Commutative Algebra代考|Basic equivalences

(iii)如果$M \hookrightarrow N$，则自然图$\operatorname{Hom}(N, E) \rightarrow \operatorname{Hom}(M, E)$是满射的。
(iv)逆变函子$\operatorname{Hom}(, E)$是精确的。
(v) r模的任何短的精确序列
$$0 \rightarrow E \stackrel{\iota}{\rightarrow} M \rightarrow N \rightarrow 0$$

## 数学代写|交换代数代写Commutative Algebra代考|Baer’s Criterion

(i) $E$是单射的。
(ii)对于$R$的每个理想非零$I$，每个$R$ -module映射$\varphi: I \rightarrow E$扩展为$R$ -module映射$\Phi: R \rightarrow E$。

(ii) $\Longrightarrow$ (1):让 $M$ 做一个 $R$的子模块 $N$ 和 $\varphi: M \rightarrow E$ 一个 $R$-module map。我们需要证明这一点 $\varphi$ 可扩展至 $N$． 现在是集合 $\mathcal{P}$ 成对的 $\left(N^{\prime}, \varphi^{\prime}\right)$ 有 $M \subset N^{\prime} \subset N$ 和 $\varphi: N^{\prime} \rightarrow E$ 地图扩展 $\varphi$ 是非空的，并且有明显的偏序，对于其中的任何元素链的并集 $\mathcal{P}$ 又是的元素吗 $\mathcal{P}$． 根据佐恩引理，存在一个极大元素 $\varphi^{\prime}: N^{\prime} \rightarrow E$． 我们的任务就是证明这一点 $N^{\prime}=N$．

$$I=\left(N^{\prime}: x\right)=\left{r \in R \mid r x \subset N^{\prime}\right}$$

$$I \stackrel{\cdot x}{\rightarrow} N^{\prime} \stackrel{\varphi}{\rightarrow} E ;$$

$$\varphi^{\prime \prime}\left(x^{\prime}+r x\right)=\varphi^{\prime}\left(x^{\prime}\right)+\psi(r) .$$

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